 In this video I want to talk about linear impact. So linear impact is when we apply a force quickly to a rigid body and we're doing so in a linear sense rather than say rotational. So when we need an example to talk through this probably the easiest one is to say we have a weight and we drop it from some height h and it's going to land on a spring which will represent our rigid body in this case. Once it makes contact with that spring of course then it's going to compress the spring by some amount delta until the whole system comes to arrest at some new compressed position you know where delta is a maximum that it's going to reach. The spring has a spring constant k and we have this the mass has a weight w and so forth. So we can start to analyze the system by looking at the energy of what's happening right. First we can say that there's potential energy change right as this weight moves down it travels through a distance h and also a distance delta till it you know reaches a new at rest position so it has a change in energy of that potential energy and its weight force times distance so weight times h plus delta. So this is that change in potential energy energy that we observe. Now you know we could also consider the kinetic energy which is you know the the speed at which it's traveling however we can kind of bypass that if we say it starts from rest at some height and ends at rest when the spring is compressed so we can kind of bypass that that moving state and just go between the two stationary states. Now the change then is we're storing energy in that spring so we need the the stored energy within the spring which we have an equation for one half k delta squared. So this is the the energy that we store within the spring that that change of potential energy has been transferred into the spring so we can equate those two things. Now there's a couple things I want to define before we move on too far from this I'm just going to do that over here one is I want to define what we'll call an equivalent force and this equivalent force is going to be equal to k delta and basically this is the the equivalent force that we would apply to the spring in a static loading scenario that would depress the spring by the same amount delta as the the falling box would. So taking that that energy of the box or the weight and saying well let's pretend it was static we'll give that what we'll call the equivalent force and then another thing that our relationship that I want to spell out is that we can equate the weight to something we'll call the static deflection. So again taking this this scenario that suppose we took that that box and just set it nicely slowly on the spring it would deform the spring by a certain amount delta which we'll call delta st for delta static which would presumably be less than delta right because the box is not moving with the velocity we're just setting its weight on there. So we kind of have these two scenarios one we're equating the moving weight w with an equivalent force based on how much it it deflects the spring while it's moving at a velocity and then we have this static deflection which is if we just set the box nicely on the spring how much we would expect it to deform as a result and that'll become useful as I work through these these other equations. So if I go back then and take my my energy balance I can substitute in from this second relationship k you know is going to be equal to w over delta st so I can substitute in and get one half delta squared over delta delta st times w as a relationship and if you look at this equation and think about it in terms of delta which is really the the the changing or unknown variable in this case the one we can't just you know we can't just get directly then we see that it's in a quadratic form right there's a delta on the left hand side and a delta squared on the right hand side so without walking through all the math we can solve that quadratic equation and what we get is delta is equal to delta st times one plus the square root of one plus two h over delta st and these are all static quantities so this is a useful equation for us and it gives us the ability to determine you know based on how high the box is and what its static deformation delta st would be you know what we can expect to to deform the spring due to that motion we can also you know based on this equation over here fe equals k delta we can rewrite this in terms of fe so get that equivalent force which basically turns a dynamic problem into a static problem we can rewrite this equation and get w one plus square root one plus two h over delta st and great now we have a relationship between the static and dynamic scenarios and you if you look at this first equation you see delta on the left is dynamic delta st on the right is static and then we have this thing in parentheses and for the force equation we have our static equivalent so fe and uh which represents a dynamic scenario and then we have the weight which is presumably unknown value and what we find then is that we have this quantity which is common between these two equations which is the the quantity in parentheses here and we call this our impact factor so effectively what that means is that it's a magnification a multiple of the static scenario w or delta st which gives us the dynamic scenario delta or force equivalent so it's it's kind of a multiplier which gives us or allows us to account for that scenario now sometimes we may not know the height from which something's falling but we may know a velocity so a good example of this would be like you're swinging a hammer and then you you strike something the the height from which the hammer is falling doesn't really make sense right that's that's not really a thing in that regard but we could know the velocity of the hammer right before it strikes the rigid body so maybe we want to formulate our equations in terms of velocity rather than change in height well again if we go back to basics of of energy so I'm just going to kind of separate my window here if we go back to the basics of energy we can write that v squared equals 2 gh so this is just equating uh mgh and one half mv squared you know potential and kinetic energy and then and then striking out you know the common terms and and solving for v squared if we do that then we can rewrite our equations as this one plus square root one plus v squared over g delta st and same for force equivalent one plus square root one plus v squared over g delta st so this gives us a slightly different formulation all it really changed is the the values that go into our impact factor to be in terms of velocity rather than you know height off of our off of our object so useful in some scenarios there's a couple of simplifications or or special scenarios for this that we might want to consider one is that suppose we have something which we might call a suddenly applied load oops that's not at all anything so a suddenly applied load would be you know if you go back to that that picture on the left of the screen we have this this weight um dropped from a distance suppose we reduced h until it was zero and we basically are just hovering the box right over the spring and then we let it go so there's no change in height there's no initial velocity and and all of those terms drop to zero well then basically what we end up having as you can see from our impact factor equation if v is zero then this term goes to zero we have square root of one which is one one plus one is two so all of that really is just to say that in this special case of a suddenly applied load the impact factor is equal to two so that can give us a sense of of what we might expect to find um for for what this impact factor would typically be so in a suddenly applied load we have an impact factor of two so no initial velocity no initial height we just sort of release the weight onto our thing or or release the object okay scroll down to give me some some white space now we can simplify this equation a little further which allows us to do some more things uh often we would expect h to be much much larger than delta st so if we drop something from a from a fairly significant height our and our spring is sufficiently stiff you know it doesn't have to be a spring it can be literally like a table or something is sufficiently stiff we would expect that deformation to be very small right much much smaller than h and if we have a large number next to a small number you might recognize from some math courses that that usually means we can cancel some things out because they're effectively negligible right so if we do that we can say delta st then is equal to delta or excuse me delta is equal to delta st square root two h over delta st so i've canceled out some terms based on that being small and i'm going to do some mathematical um reorganization here basically if i pulled the delta st out um you know divided by the delta st i have a power of one power of one half in the denominator um and what i end up with is i can move that delta st inside my square root and using my substitution from before i have delta st v squared over g as another formulation of this equation and i can substitute or rewrite v squared as kinetic energy so two i'm putting these squiggly lines around kinetic energy so we don't confuse things in a minute here so two times kinetic energy over spring stiffness k making some additional substitutions so now i have an equation in terms of energy which might be useful and doing the same thing for my equivalent force i can write two h over delta st again doing some rearranging and substitution because delta st is equal to w over k that's an important thing to remember and if i substitute that in i can write this equation as two w h k i could write this as v squared k w over g which then gives me a similar formulation in terms of energy of two k e times spring constant k so now i have a formulation in terms of kinetic energy two times kinetic energy times k it gives me that that force equivalent okay now probably the whole reason for doing any of this is to figure out what my stress impact on my stress is so if i want to know what's going on with stress due to this impact loading i can take my normal stress equation but now i'm using my equivalent force over a unit area a good way to think about this problem is suppose i have a bolt or something pinned and it has a mass around it and that mass is falling and it's going to strike this thing that's holding it up and i have a cross sectional area in here great i can apply and figure out what my stress is using this normal equation and if i do some rearranging based on my calculations that i have up here then i can rewrite this equation as square root two times kinetic energy times young's modulus over a l so cross sectional area times length you'll see why i put these squiggly lines around kinetic energy otherwise it would get confused with young's modulus e and this of course would also be equivalent to two kinetic energy young's modulus a l is just volume so i can look at how that stress is distributed through the volume of of my part great so now i have a formulation for stress based on kinetic energy um you know velocity of my thing one half mv squared would be kinetic energy and young's modulus which is a material property and volume which is just representing my geometry and if i rearrange and solve this equation i get another quantity basically just solving for kinetic energy i get stress squared volume over two times young's modulus and this quantity is called the impact energy capacity and really what it's doing is it's saying if i have a limit for my stress so if i take sigma and i substitute in my materials stress limit i can calculate then how much energy my rigid body can absorb so what is its capacity for absorbing that load energy all right so i'm going to go ahead and stop there